Question

A curve is given between the potential energy \(U\) of a particle and its position on the \(x\)-axis as shown. Given: \[ \tan\theta_1 = 1,\qquad \tan\theta_2 = 3,\qquad \tan\theta_3 = -\frac{1}{2} \] If \(F_{AB}\) is the force acting on the particle during motion from \(A\) to \(B\), similarly \(F_{BC}\), \(F_{CD}\) and \(F_{DE}\) are the forces during \(B\) to \(C\), \(C\) to \(D\) and \(D\) to \(E\) respectively, arrange the magnitudes of these forces in decreasing order.

• A. \(F_{BC}>F_{AB}>F_{CD}>F_{DE}\)
• B. \(F_{BC}>F_{AB}>F_{DE}>F_{CD}\)
• C. \(F_{AB}>F_{BC}>F_{DE}>F_{CD}\)
• D. \(F_{BC}>F_{DE}>F_{AB}>F_{CD}\)

Hint:

For potential energy graphs:
Force is given by the negative slope of the \(U\)--\(x\) graph
Steeper slope means larger force
Flat regions correspond to zero force

Answer 1

The Correct Option is B

Force acting on a particle is related to the potential energy by: \[ F = \left| -\frac{dU}{dx} \right| \] Hence, the magnitude of force is equal to the magnitude of the slope of the \(U\)--\(x\) graph.
On segment \(AB\): slope \(= \tan\theta_1 = 1 \Rightarrow F_{AB} \propto 1\)
On segment \(BC\): slope \(= \tan\theta_2 = 3 \Rightarrow F_{BC} \propto 3\)
On segment \(CD\): slope \(= 0 \Rightarrow F_{CD} = 0\)
On segment \(DE\): slope \(= |\tan\theta_3| = \frac{1}{2} \Rightarrow F_{DE} \propto \frac{1}{2}\) Arranging in decreasing order of magnitude: \[ F_{BC}>F_{AB}>F_{DE}>F_{CD} \]